Skip to main content

Section 10.4 Completeness

Given a vector and an orthonormal basis, it is easy to determine the components of the vector in the given basis. For example, if

\begin{equation} \FF = F_x \,\xhat + F_y \,\yhat + F_z \,\zhat\tag{10.4.1} \end{equation}

then dotting both sides of this equation with \(\xhat\) yields the \(x\)-component of the vector, namely \(F_x\text{.}\)

\begin{align} \xhat\cdot\FF \amp= \xhat\cdot(F_x \,\xhat + F_y \,\yhat + F_z \,\zhat)\notag\\ \amp= F_x\tag{10.4.2} \end{align}

where the orthonormality of the basis vectors has been used in the last step.

Put differently,

\begin{equation} \FF = (\FF\cdot\xhat)\,\xhat + (\FF\cdot\yhat)\,\yhat + (\FF\cdot\zhat)\,\zhat\text{.}\tag{10.4.3} \end{equation}

All we need to make this idea work is an orthogonal basis. If the basis is not normalized, the method still works, but we will need to account for the normalization of the basis vectors explicitly.

In Section 10.3, the “dot product”, that is, the inner product, is the integral from \(0\) to \(L\text{,}\) and the orthogonal basis elements are the trigonemtric functions \(\left\{1,\cos\left(\frac{2\pi m x}{L}\right),% \sin\left(\frac{2\pi m x}{L}\right)\right\}\) with \(m\) a positive integer. There are an infinite number of basis vectors! But what vector space do they span?

It turns out that the any periodic function that is reasonably well behaved (for instance, square integrable) can be expanded uniquely in terms of this basis. The technical language for this concept is that this basis is complete on the vector space of (square-integrable) periodic functions with period \(L\text{.}\)